[mage lang="" source="flickr"]stewart calculus book answers[/mage]
Calculus homework help!?

This is a problem from Calculus Early Transcendentals, James Stewart, 6 Edition. The section is 4.1 and problem 59. You are trying to find absolute extrema and the problem is f(x)=xe^((x^2)/8), and the answer in the back of the book is 2 (the interval is bounded from -1 to 4) and 1. and i cannot figure out where the 1 came from.

f(x) = xe^(x²/8)

Extrema occur where f’(x) = 0, so find all roots of f’(x) between -1 and 4.

f’(x) = ¼x²e^(x²/8) + e^(x²/8)

f’(x) = (¼x² + 1)(e^(x²/8))

e^(x²/8) is an exponential, so it is never equal to zero. So f’(x) is 0 when (¼x² + 1) = 0.

(¼x² + 1) = 0
¼x² = -1
x² = -4
x = 2i, x = -2i

There are no real zeros to f’(x), so f(x) has no natural extrema. The extrema are only present because the interval is restricted. So one extrema will be at f(-1) and the other at f(4).

f(-1) = -e^(1/8) = -1.13 —–> Minimum
f(4) = 4e² = 29.6 ——> Maximum

I don’t know how your book got that answer, but nothing of interest happens at y = 2, or y = 1. f(x) itself has a root and switches concavity at x = 0, but that’s about the only thing going on with this curve.

Black Purdue documentary film

You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
3 Responses
1. stab says:

To guard your Facebook Login on public computer systems keep away from clicking the maintain me logged in box when logging into your Fb account. While you

2. Fantastic beat ! I wish to apprentice while you amend your site, how could i subscribe for
a weblog website? The account aided me a acceptable deal.
I were a little bit familiar of this your broadcast offered bright
transparent concept

3. Bing says:

Very shortly this site will be famous among all blog users,
due to it’s pleasant articles

Visit my website :: Bing